Chapter 18. Kinetics of Rigid Bodies in Three Dimensions

Introduction

Rigid Body Angular Momentum in Three Dimensions

Principle of Impulse and Momentum

Kinetic Energy

Motion of a Rigid Body in Three Dimensions

Euler’s Equations of Motion and D’Alembert’s Principle

Motion About a Fixed Point or a Fixed Axis

Motion of a Gyroscope. Eulerian Angles

Steady Precession of a Gyroscope

Motion of an Axisymmetrical Body Under No Force

Three dimensional analyses are needed to determine the forces and moments

on the gimbals of gyroscopes, the joints of robotic welders, and the supports

of radio telescopes.

Introduction

• The fundamental relations developed for the

plane motion of rigid bodies may also be applied

to the general motion of three dimensional bodies.

• The relation which was used

to determine the angular momentum of a

rigid slab is not valid for general three

18.1 dimensional bodies and motion.

The current chapter is concerned with

evaluation of the angular momentum and

its rate of change for three dimensional

motion and application to effective forces,

the impulse-momentum and the work-

energy principles.

ω

IHG =

GG HM

amF

=

=

∑∑

18.1A Angular Momentum of a Rigid Body in Three Dimensions

Angular momentum of a body about its mass center,

The x component of the angular momentum,

( ) ( )[ ]∑∑==

′××′=×′=n

iiii

n

iiiiG mrrmvrH

11Δ

ω∆

( ) ( )[ ]

( ) ( )[ ]

( ) ∑∑∑

∑

∑

===

=

=

−−+=

−−−=

′×−′×=

n

iiiiz

n

iiiiy

n

iiiix

n

iiixiziiyixi

n

iiyiiziix

mxzmyxmzy

mzxzxyy

mrzryH

111

22

1

1

ΔΔΔ

Δ

Δ

ωωω

ωωωω

ωω

( )2 2x x y zH y z dm xy dm zx dmω ω ω= + − −∫ ∫ ∫

x x xy y xz zI I Iω ω ω= + − −

y yx x y y yz z

z zx x zy y z z

H I I I

H I I I

ω ω ω

ω ω ω

= − + −

= − − +

Transformation of into is characterized by the inertia tensor for the body,

• With respect to the principal axes of inertia,

The angular momentum of a rigid body and its

angular velocity have the same direction if, and only if,

is directed along a principal axis of inertia.

ω

GH

ω

ω

+−−−+−−−+

zzyzx

yzyyx

xzxyx

IIIIIIIII

GH

′

′

′

z

y

x

II

I

000000

x x x y y y z z zH I H I H Iω ω ω′ ′ ′ ′ ′ ′ ′ ′ ′= = =

The momenta of the particles of a rigid body can be reduced

to:

• The angular momentum about any other given point O is

vmL

== momentumlinear

GHG about momentumangular =

zzyzyxzxz

zyzyyxyxy

zxzyxyxxx

IIIH

IIIH

IIIH

ωωω

ωωω

ωωω

+−−=

−+−=

−−+=

GO HvmrH

+×=

The angular momentum of a body constrained to rotate about a

fixed point may be calculated from

• Or, the angular momentum may be

computed directly from the moments

and products of inertia with respect to the Oxyz frame.

( )

( )[ ]∑

∑

=

=

××=

×=

n

iiii

n

iiiO

mrr

mvrH

1

1

Δ

Δ

ω

GO HvmrH

+×=

x x x xy y xz z

y yx x y y yz z

z zx x zy y z z

H I I IH I I IH I I I

ω ω ω

ω ω ω

ω ω ω

= + − −

= − + −

= − − +

18.1B Principle of Impulse and Momentum

• The principle of impulse and momentum can be applied directly to the three-dimensional

motion of a rigid body,

Syst Momenta1 + Syst Ext Imp1-2 = Syst Momenta2

• The free-body diagram equation is used to develop component and moment equations.

• For bodies rotating about a fixed point, eliminate the impulse of the reactions at O by

writing equation for moments of momenta and impulses about O.

18.1C Kinetic Energy

Kinetic energy of particles forming rigid body,

• If the axes correspond instantaneously with the

principle axes,

• With these results, the principles of work and energy and conservation of energy may be

applied to the three-dimensional motion of a rigid body.

)22

2(

Δ

Δ

222212

21

1

2212

21

1

2212

21

xzzxzyyz

yxxyzzyyxx

n

iii

n

iii

II

IIIIvm

mrvm

vmvmT

ωωωω

ωωωωω

ω

−−

−+++=

′×+=

′+=

∑

∑

=

=

)( 222212

21

zzyyxx IIIvmT ′′′′′′ +++= ωωω

• Kinetic energy of a rigid body with a fixed point,

• If the axes Oxyz correspond instantaneously with the principle

axes Ox’y’z’,

)22

2( 22221

xzzxzyyz

yxxyzzyyxx

II

IIIIT

ωωωω

ωωωωω

−−

−++=

)( 22221

zzyyxx IIIT ′′′′′′ ++= ωωω

Sample Problem 18.1

Rectangular plate of mass m that is suspended from two

wires is hit at D in a direction perpendicular to the plate.

Immediately after the impact, determine a) the velocity of

the mass center G, and b) the angular velocity of the plate.

STRATEGY:

• Apply the principle of impulse and momentum. Since the initial momenta is zero, the

system of impulses must be equivalent to the final system of momenta.

• Assume that the supporting cables remain taut such that the vertical velocity and the

rotation about an axis normal to the plate is zero.

• Principle of impulse and momentum yields to two equations for linear momentum and

two equations for angular momentum.

• Solve for the two horizontal components of the linear and angular velocity vectors.

MODELING and ANALYSIS:

• Apply the principle of impulse and momentum. Since the initial momenta is zero, the

system of impulses must be equivalent to the final system of momenta.

• Assume that the supporting cables remain taut such that the vertical velocity and the

rotation about an axis normal to the plate is zero.

Since the x, y, and z axes are principal axes of inertia,

kvivv zx

+= ji yx

ωωω +=

jmaimbjIiIH yxyyxxG

ωωωω 21212

121 +=+=

• Principle of impulse and momentum yields two equations for linear momentum and two

equations for angular momentum.

• Solve for the two horizontal components of the linear and angular velocity vectors.

0 xmv=0xv =

Δ zF t mv− =Δzv F t m= −

12

2112

Δ x

x

bF t H

mb ω

=

=

12

2112

Δ y

y

aF t H

ma ω

− =

=

( )Δv F t m k= −

6 Δx F t mbω = ( )6 Δy F t maω = −

( )6 ΔF t ai b jmab

ω = +

( )kmtFv

Δ−

2 21 112 12G x yH mb i ma jω ω= +

( )jbiamab

tF

+=Δ6ω

REFLECT and THINK:

• Equating the y components

of the impulses and

momenta and their

moments about the z axis,

you can obtain two

additional equations that

yield TA = TB = ½W.

• This verifies that the wires

remain taut and that the

initial assumption was

correct. If the impulse were

at G, this would reduce to a

two-dimensional problem.

Sample Problem 18.2

A homogeneous disk of mass m is mounted on an axle

OG of negligible mass. The disk rotates counter-clockwise

at the rate w1 about OG.

Determine: a) the angular velocity of the disk, b) its

angular momentum about O, c) its kinetic energy, and d)

the vector and couple at G equivalent to the momenta of

the particles of the disk.

STRATEGY:

• The disk rotates about the vertical axis through O as well as about OG. Combine the

rotation components for the angular velocity of the disk.

• Compute the angular momentum of the disk using principle axes of inertia and noting

that O is a fixed point.

• The kinetic energy is computed from the angular velocity and moments of inertia.

• The vector and couple at G are also computed from the angular velocity and moments

of inertia.

MODELING and ANALYSIS:

• The disk rotates about the vertical axis through O as well as

about OG. Combine the rotation components for the angular

velocity of the disk.

Noting that the velocity at C is zero,

ji

21 ωωω +=

( ) ( )( )

LrkrL

jriLjirv CC

12

12

2100

ωωωω

ωωω

=−=

−×+=

=×=

( )1 1i r L jω ω ω= −

• Compute the angular momentum of the disk using principle

axes of inertia and noting that O is a fixed point.

• The kinetic energy is computed from the angular velocity and moments of inertia.

kIjIiIH zzyyxxO

ωωω ++=

( ) jLri

11 ωωω −=

( )( )( )( )

2112

2 2114

2 214 0 0

x x x

y y y

z z z

H I mr

H I mL mr r L

H I mL mr

ω ω

ω ω

ω

= =

= = + −

= = + =

( )( )2 2 21 11 12 4OH mr i m L r r L jω ω= − +

22 21

18 26 rT mrL

ω

= +

( )( )( )[ ]2

12

4122

12

21

22221

LrrLmmr

IIIT zzyyxx

ωω

ωωω

−++=

++=

The vector and couple at G are also computed from

the angular velocity and moments of inertia.

REFLECT and THINK:

• If the mass of the axle were not negligible and it was instead

modeled as a slender rod with a mass Maxle, it would also contribute

to the kinetic energy

Taxle = ½(1/3 MaxleL2) ω22 and the momenta

Haxle = ½(1/3 MaxleL2) ω2 of the system.

kmrvm

1ω=

( ) jLri

11 ωωω −=

( )2 21 112 4

G x x y y z zH I i I j I k

mr i mr r L j

ω ω ω

ω ω′ ′ ′= + +

= + −

2112 2G

rH mr i jL

ω = −

18.2 Motion of a Rigid Body in Three Dimensions

18.2A Rate of Change of Angular Momentum

Angular momentum and its rate of change are taken with

respect to centroidal axes GX’Y’Z’ of fixed orientation.

Transformation of into is independent of the

system of coordinate axes.

Convenient to use body fixed axes Gxyz where moments

and products of inertia are not time dependent.

• Define rate of change of change of with respect

to the rotating frame,

ω GH

GH

GHM

amF

=

=

∑∑

Then,

18.2B Euler’s Eqs of Motion

With and Gxyz chosen to correspond to the

principal axes of inertia,

Euler’s Equations:

ωΩ

=

( ) GGxyzGG HHM

×+=∑ Ω

( ) kHjHiHH zyxGxyzG

++=

( ) ωΩΩ

=×+= GGxyzGG HHH

• System of external forces and effective forces are equivalent for general

three dimensional motion.

• System of external forces are equivalent to the vector and couple,

( )( )( ) yxyxzzz

xzxzyyy

zyzyxxx

IIIM

IIIM

IIIM

ωωω

ωωω

ωωω

−−=

−−=

−−=

∑∑∑

. and GHam

18.2C,D Motion About a Fixed Point or a Fixed Axis

For a rigid body rotation around a fixed point,

For a rigid body rotation around a fixed axis,

x xzH I ω= − y yzH I ω= − z zH I ω=

( ) OOxyzO

OO

HH

HM

×+=

=∑Ω

( )( )

( )( ) ( ) 2

O O OOxyz

xz yz z

xz yz z

xz yz z xz yz

M H H

I i I j I k

k I i I j I k

I i I j I k I j I i

ω

ω

ω ω

α ω

= + ×

= − − +

+ × − − +

= − − + + − +

∑

2

2

x xz yz

y yz xz

z z

M I I

M I I

M I

α ω

α ω

α

= − +

= − +

=

∑∑∑

If symmetrical with respect to the xy plane,

• If not symmetrical, the sum of external moments will not be zero, even if α

= 0,

• A rotating shaft requires both static and dynamic balancing

to avoid excessive vibration and bearing reactions.

αzzyx IMMM === ∑∑∑ 00

022 === ∑∑∑ zxzyyzx MIMIM ωω

( )0=ω ( )0≠ω

Sample Problem 18.3

Rod AB with mass m = 20 kg is pinned at A to a vertical axle which

rotates with constant angular velocity ω = 15 rad/s. The rod

position is maintained by a horizontal wire BC.

Determine the tension in the wire and the reaction at A.

STRATEGY:

• Evaluate the system of effective forces by reducing them to a vector attached at G

and couple

• Expressing that the system of external forces is equivalent to the system of effective

forces, write vector expressions for the sum of moments about A and the summation of

forces.

• Solve for the wire tension and the reactions at A.

am

.GH

MODELING and ANALYSIS:

• Evaluate the system of effective forces by reducing them to

a vector attached at G and couple

am .GH

G x x y y z zH I i I j I kω ω ω= + +

2 21 112 120

cos sin 0x y z

x y z

I mL I I mLω ω β ω ω β ω

= = == − = =21

12 cosGH mL iω β= −

Expressing that the system of external forces is equivalent to

the system of effective forces, write vector expressions for

the sum of moments about A and the summation of forces.

( )effAA MM ∑∑ =

( ) ( ) ( )( ) ( )

1.732 0.5 196.2 0.866 2250 649.5

1.732 98.1 1948.5 649.5

J TI I J J I K

T K K

× − + × − = × − +

− = +

1557 NT =( )

effF F=∑ ∑

1613 196.2 2250X Y ZA I A J A K I J I+ + − − = −

( ) ( )697 N 196.2 NA I J= − +

•

Gyroscopes are used in the navigation system of the Hubble telescope, and can

also be used as sensors . Modern gyroscopes can also be MEMS (Micro Electro-

Mechanical System) devices, or based on fiber optic technology.

* Motion of a Gyroscope. Eulerian Angles *

A gyroscope consists of a rotor with its mass center fixed in space

but which can spin freely about its geometric axis and assume any

orientation.

From a reference position with gimbals and a reference diameter of

the rotor aligned, the gyroscope may be brought to any orientation

through a succession of three steps:

a) rotation of outer gimbal through φ about AA’,

b) rotation of inner gimbal through θ about BB’,

c) rotation of the rotor through ψ about CC’.

• φ θ ψ : Eulerian Angles

spin of rate nutation of rate precession of rate

=

=

=

Ψθφ

The angular velocity of the gyroscope,

Equation of motion,

kjK

Ψθφω ++=

( ) OOxyzOO HHM

×+=∑ Ω

( )with sin cos

sin cos

K i k

i j k

θ θ

ω φ θ θ φ θ

= − +

= − + + Ψ +

( )sin cosOH I i I j I k

K j

φ θ θ φ θ

φ θ

′ ′= − + + Ψ +

Ω = +

( ) ( )( ) ( )( )

2

sin 2 cos cos

sin cos sin cos

cos

x

y

z

M I I

M I I

dM Idt

φ θ θφ θ θ φ θ

θ φ θ θ φ θ φ θ

φ θ

′= − + + Ψ +

′= − + Ψ +

= Ψ +

∑∑

∑

Steady Precession of a Gyroscope

Steady precession,

Couple is applied about an axis perpendicular to the precession and

spin axes

constant are ,, ψφθ

ki

kIiIH

ki

zO

z

θφθφΩ

ωθφ

ωθφω

cossin

sin

sin

+−=

+′−=

+−=

( ) jIIHM

z

OO

θφθφω

Ω

sincos′−=

×=∑

When the precession and spin axis are at a right angle,

Gyroscope will precess about an axis perpendicular to both the

spin axis and couple axis.

jIM O

φΨ

θ

=

°=

∑90

Motion of an Axisymmetrical Body Under No Force

Consider motion about its mass center of an axisymmetrical

body under no force but its own weight, e.g., projectiles,

satellites, and space craft.

Define the Z axis to be aligned with and z in a rotating

axes system along the axis of symmetry. The x axis is chosen

to lie in the Zz plane.

• θ = constant and body is in steady precession.

• Note:

constant 0 == GG HH

GH

xGx IHH ωθ ′=−= sin IHG

x ′−=

θω sin

θγωω tantan

II

z

x′

==−

0y yH I ω′= = 0yω =cosz G zH H Iθ ω= = cosG

zH

Iθω =

• Two cases of motion of an axisymmetrical body which under

no force which involve no precession:

• Body set to spin about its axis of symmetry,

and body keeps spinning

about its axis of symmetry.

• Body is set to spin about its transverse axis,

and body keeps spinning about the given transverse axis.

aligned are and0

G

zz

HH

ω

ω ==

aligned are and

0

G

xx

HH

ω

ω ==

The motion of a body about a fixed point (or its mass center) can be

represented by the motion of a body cone rolling on a space cone.

In the case of steady precession the two cones are circular.

• I < I’. Case of an elongated body. γ θ< and the vector ω

lies inside the angle ZGz. The space cone and body cone are tangent

externally; the spin and precession are both counterclockwise from

the positive z axis. The precession is said to be direct.

• I > I’. Case of a flattened body. γ θ< and the vector ω lies outside the

angle ZGz. The space cone is inside the body cone; the spin and precession

have opposite senses. The precession is said to be retrograde.